Given the parametric equations:

$x = 4t, \quad y = 4t^{-1},$

we are asked to find $\frac{dy}{dx}$.

Step 1: Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$

$\frac{dx}{dt} = \frac{d}{dt}(4t) = 4,$

$\frac{dy}{dt} = \frac{d}{dt}(4t^{-1}) = 4 \cdot (-t^{-2}) = -\frac{4}{t^2}.$

Step 2: Apply the chain rule for parametric differentiation

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\frac{4}{t^2}}{4} = -\frac{1}{t^2}.$

Step 3: Express in terms of $x$

Since $x = 4t$, we can solve for $t$:

$t = \frac{x}{4}.$

Now substitute this expression for $t$ into the equation for $\frac{dy}{dx}$:

$\frac{dy}{dx} = -\frac{1}{\left(\frac{x}{4}\right)^2} = -\frac{1}{\frac{x^2}{16}} = -\frac{16}{x^2}.$

Thus, the derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = -\frac{16}{x^2}.$

Would you like more details or have any questions about the process?

Here are 5 related questions:

1. How would you find the second derivative $\frac{d^2y}{dx^2}$ for the given parametric equations?
2. Can you derive $\frac{dx}{dy}$ from the parametric equations?
3. What would $\frac{dy}{dx}$ be if $x = 2t$ and $y = 2t^{-1}$?
4. How would the result change if $x = t^2$ and $y = t^{-2}$?
5. How can you interpret $\frac{dy}{dx}$ geometrically in parametric curves?

Tip: When differentiating parametric equations, always express the derivative in terms of the original variable $x$ for consistency.